On Fri, 18 Jan 2013, Butch Malahide wrote:> On Jan 19, 1:05 am, William Elliot <ma...@panix.com> wrote:> >> > > > Does this generalize to every uncountable limit ordinal eta,> > > > that f in C(eta,R) is eventually constant and thusly the Cech> > > > Stone compactification of of eta is eta + 1? Does eta need> > > > to have an uncountable cofinality for this generalization?> >> > > Yes, the same argument that works for omega_1 also works for any > > > ordinal of uncountable cofinality.> >> > > Let X be a linearly ordered topological space (i.e., a linearly > > > ordered set with its order topology) in which every increasing > > > sequence converges. [Examples: any ordinal of uncountable > > > cofinality; the long line; any countably compact LOTS.] Call a > > > subset of X "bounded" if it has an *upper* bound in X, "unbounded" > > > otherwise. Observe that (1) the union of countably many bounded sets > > > is bounded, and (2) the intersection of countably many unbounded > > > closed sets is unbounded.> >> > Does not (1) hold because every increasing sequence converges?> > Wouldn't every countable set has an upper bound suffice?> > Yes, "every countable set has an upper bound" would suffice for (1),> but not for (2).

Why not? I've reviewed a proof for that theorem and see than insteadof supremum or limits, that upper bounds would suffice. In fact,that every countable set has an upper bound is equivalent to uncountablecofinality.

> > > Let Y be a topological space which is hereditarily Lindelof and such> > > that, for each point y in Y, the set {y} is the intersection of> > > countably many closed neighborhoods of y. [Example: any separable> > > metric space.]> >> > > THEOREM. If X and Y are as stated above, then every function f in> > > C(X,Y) is eventually constant.> >> > > PROOF. We may assume that X has no greatest element. For S a subset of> > > Y, let g(S) = {x in X: f(x) is in S}. Let Z = {y in Y: g({y}) is> > > bounded}.> >> > Is assuming X has no greatest element, an additional premise?> > No. The assumption that X has no greatest element is made without loss> of generality, because the contrary case is trivial: if X has a> greatest element, then every function with domain X is eventually> constant.

Uncountable cofinality implies X has no max.

> > Is Z closed?> > It will be shown that Y\Z contains a single point c, the eventual> value of the function f, and so Z = Y\{c}. To ask whether Z is closed> is to ask whether c is an isolated point of Y. It could, but it> doesn't have to be.> > > > CLAIM. Each point z in Z has a neighborhood V_z such that g(V_z) is> > > bounded.> >> > > PROOF OF CLAIM. Let {U_n: n in N} be a countable collection of closed> > > neighborhoods of z whose intersection is {z}. Assume for a> > > contradiction that each set g(U_n) is unbounded. Since f is> > > continuous, each g(U_n) is an unbounded closed set. By property (2)> > > above, g({z}) = /\{g(U_n): n in N} is unbounded, contradicting the> > > assumption that z is in Z.> >> > > Thus the set Z is covered by open sets V_z such that g(V_z) is> > > bounded. Since Y is hereditarily Lindelof, it follows that Z is> > > covered by countably many open sets V such that g(V) is bounded. In> > > view of property (1) above, it follows that g(Z) is bounded. Since> > > g(Y) = X is unbounded, Y\Z is nonempty. All we have left to show is> > > that Y\Z consists of a single point. Assume for a contradiction that Y> > > \Z contains two distinct points c and d. Thus g({c}) and g({d}) are> > > unbounded closed subsets of X. By property (2) above, the intersection> > > of g({c}) and g({d}) is unbounded; in particular, it is nonempty. Let> > > x be a point in the intersection of g({c}) and g({d}). Then c = f(x) => > > d, contradicting the assumption that c and d are two distinct points.>